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If I have a function $f: (-a,a) \rightarrow \mathbb{R}$ and $A \in \mathbb{R}$. Then how can I prove that $$ \lim_{x \to 0} f(x) = A \Leftrightarrow \lim_{x \to 0 } f(\sin{x}) =A $$ using real analysis methods?

I'm thinking that I should use the property that $\sin{x}$ is bounded. But I don't know where to start.

Benjamin
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Hannah
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  • Can you utilize the fact that $\lim_{x\to 0}\sin x = 0$? – MSDG Apr 10 '19 at 19:31
  • yes, we can make use of that. – Hannah Apr 10 '19 at 19:33
  • Use the fact that $$\Big(\lim_{x\rightarrow x_0}g(x) =y_0 \Big) \Rightarrow \Big(\lim_{x\rightarrow x_0} f(g(x)) = \lim_{y\rightarrow y_0} f(y)\Big) $$ – Adam Latosiński Apr 10 '19 at 20:48
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    @AdamLatosiński That requires assuming that $f$ is continuous at $x=x_0$, but in this problem we're not given that $f$ is continuous at 0. For example, if $f$ is discontinuous at 0 then $\lim_{x\to 0} f(x) = A$ does not imply $\lim_{x\to 0} f(x \sin(1/x)) = A$ even though $\lim_{x\to 0} x \sin(1/x) = 0$. So, the solution is going to need to use something else about $\sin x$, e.g. the fact that it's nonzero on a punctured neighborhood of 0. – Daniel Schepler Apr 10 '19 at 21:48
  • Its due to rule of substitution for limits. See this answer: https://math.stackexchange.com/a/1073047/72031 The result in question depends on the fact that $\sin x$ is invertible in some deleted neighborhood of $0$ and we have $\lim_{x\to 0}\sin x=0=\lim_{x\to 0}\arcsin x$. – Paramanand Singh Apr 11 '19 at 14:43
  • @DanielSchepler: the result holds provided $g(x) \neq y_0$ in some deleted neighborhood of $x_0$. The continuity of $f$ is not needed in this more general scenario. – Paramanand Singh Apr 11 '19 at 14:44

3 Answers3

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If $\lim_{x\to0}f(x)=A$, then, for all $\varepsilon>0$, there exists a $\delta$ such that $|x-0|<\delta\Rightarrow|f(x)-A|<\varepsilon$ by definition.

Let $x=\sin u$. Then, since $|\sin u|<|u|$ for nonzero $u$, we have $$|u-0|<\delta\Rightarrow|\sin u-0|<\delta\Rightarrow |f(\sin u)-A|<\varepsilon$$ so $\lim f(x)=A \Rightarrow \lim f(\sin x)=A$. To see that the reverse implication is true, let $x=\sin^{-1}u$ and follow similar logic to show $\lim f(\sin x)=A \Rightarrow \lim f(x)=A$. This completes the proof. NB: this doesn't presuppose that either limit actually exists. If $f$ were discontinuous, neither side would have a limit.

Jam
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$\sin 0 = 0$ and $\sin$ is continuous.

So for every $\epsilon > 0$ there is a $\delta> 0$ so the if $|x-0| < \delta$ then $|\sin x - \sin 0| = |\sin x - 0| < \epsilon$.

And we are told $\lim_{x\to 0}f(x) = A$ which means for every $\epsilon > 0$ the is a $\delta > 0$ so that if $|x -0| < \delta$ then $|f(x) - A| < \epsilon$.

So for any $\epsilon > 0$ let $\gamma=\delta_1$ be the number so that $|x - 0|< \delta_1 \implies |f(x) - A| < \epsilon$.

Then let $\epsilon_2 = \gamma = \delta_1$ and let $\delta> $ be the number so that if $|x - 0| < \delta \implies |\sin x - 0| < \epsilon_2 = \gamma = \delta_1$.

So if $|x-0| < \delta$ then $|\sin x - 0| < \epsilon_2 =\delta_1 $ and $|f(\sin x) - A| < \epsilon$.

So $|x-0| < \delta \implies |f(\sin x) - A| < \epsilon$.

So $\lim_{x\to 0}f(\sin x) = A$.

fleablood
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Hint:

In some neighborhood of $x=0$, this bracketing holds:

$$\frac{2x}\pi\le\sin x\le x.$$

So applying the $\delta-\epsilon$ argument, $x$ and $\sin x$ will be interchangeable.